B. The circuit at right is made from an ideal battery 9 V connected to 4 ideal resistors as shown. i. What is the total resistance of this circuit? Show your work. R4 = 352 R1= 152 R3 = 452 ii. Find the current through each of the 4 resistors. Show your work. R2= 752 iii. Rank the potential difference across the 4 resistors from greatest to least. Explain your reasoning. iv. If resistor R3 is cut out of the circuit, without replacing it with anything, will the current through resistor R2 increase, decrease, or stay the same? Explain your answer or show your work.

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Please answer all the sections for question B (aka i,ii,iii,iv)

### Educational Website Content: Analyzing Electrical Circuits with Resistors

---

#### Circuit Analysis Problem

#### The Circuit

Attached is a circuit diagram made from an ideal battery connected to 4 ideal resistors. The diagram is detailed as follows:

- **Voltage Source (Battery):** 9 V
- **Resistor R1:** 15Ω
- **Resistor R2:** 75Ω
- **Resistor R3:** 45Ω
- **Resistor R4:** 35Ω

#### Questions and Solutions

**i. What is the total resistance of this circuit? Show your work.**

To determine the total resistance, we need to identify how resistors are connected (series or parallel) and then apply the appropriate formulas. 

- **Series Connection:** Resistors R3 (45Ω) and R2 (75Ω) are in series. 
  - Total series resistance, Rs = R3 + R2 = 45Ω + 75Ω = 120Ω

- **Parallel Connection:** The combined resistor Rs (120Ω) is in parallel with R4 (35Ω).
  - Total resistance of parallel combination, Rp = 1 / (1/Rs + 1/R4) = 1 / (1/120Ω + 1/35Ω)
  - Rp = 1 / (0.00833 + 0.02857) = 1 / 0.0369 = 27.1Ω (approx.)

- **Series Connection with R1:** Rp (27.1Ω) is in series with R1 (15Ω).
  - Total resistance, Rt = Rp + R1 = 27.1Ω + 15Ω = 42.1Ω (approx.)

**ii. Find the current through each of the four resistors. Show your work.**

The current through the entire series circuit can be found using Ohm’s law, \( I = \frac{V}{R} \).

- Total circuit resistance Rt = 42.1Ω
- Voltage V = 9V

So, \( I = \frac{9V}{42.1Ω} ≈ 0.214 A \)

- **Current through R1:** \( I_{R1} = 0.214 A \)
- **Voltage across Rp:** V(Rp) = I * Rp = 0.214A * 27.1Ω
Transcribed Image Text:### Educational Website Content: Analyzing Electrical Circuits with Resistors --- #### Circuit Analysis Problem #### The Circuit Attached is a circuit diagram made from an ideal battery connected to 4 ideal resistors. The diagram is detailed as follows: - **Voltage Source (Battery):** 9 V - **Resistor R1:** 15Ω - **Resistor R2:** 75Ω - **Resistor R3:** 45Ω - **Resistor R4:** 35Ω #### Questions and Solutions **i. What is the total resistance of this circuit? Show your work.** To determine the total resistance, we need to identify how resistors are connected (series or parallel) and then apply the appropriate formulas. - **Series Connection:** Resistors R3 (45Ω) and R2 (75Ω) are in series. - Total series resistance, Rs = R3 + R2 = 45Ω + 75Ω = 120Ω - **Parallel Connection:** The combined resistor Rs (120Ω) is in parallel with R4 (35Ω). - Total resistance of parallel combination, Rp = 1 / (1/Rs + 1/R4) = 1 / (1/120Ω + 1/35Ω) - Rp = 1 / (0.00833 + 0.02857) = 1 / 0.0369 = 27.1Ω (approx.) - **Series Connection with R1:** Rp (27.1Ω) is in series with R1 (15Ω). - Total resistance, Rt = Rp + R1 = 27.1Ω + 15Ω = 42.1Ω (approx.) **ii. Find the current through each of the four resistors. Show your work.** The current through the entire series circuit can be found using Ohm’s law, \( I = \frac{V}{R} \). - Total circuit resistance Rt = 42.1Ω - Voltage V = 9V So, \( I = \frac{9V}{42.1Ω} ≈ 0.214 A \) - **Current through R1:** \( I_{R1} = 0.214 A \) - **Voltage across Rp:** V(Rp) = I * Rp = 0.214A * 27.1Ω
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